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chubhunter [2.5K]

3 years ago

11

Real estate ads suggest that 56 % of homes for sale have garages, 27 % have swimming pools, and 16 % have both features. a) W

hat is the probability that a home for sale has a garage, but not a pool? b) If a home for sale has a garage, what's the probability that it has a pool, too? c) Are having a garage and having a pool independent events? Explain. d) Are having a garage and having a pool mutually exclusive? Explain.

1 answer:

Marizza181 [45]3 years ago

5 0

Answer:

a) 0.4

b) 0.28

c) The events are not independent

d) The events are not mutually exclusive

Step-by-step explanation:

Hi!

Lets call:

Gar = {homes with garage}

Pool = {homes with pool},

A = {homes with pool and garage} = Gar ∩ Pool

The data we are given is:

P(Gar) = 0.56

P(Pool) = 0.27

P(A) = 0.16

a) B = {homes with garage but not pool}. This set B is the set Gar without set A: B = Gar / A

P(B) = P(Gar) - P(A) = 0.4

b) This is a conditional probability:

P(Pool | Gar) = P(Gar ∩ Pool) / P(Gar) = 0.16/0.56 = 0.28

c) To be independent events, it must be, by definition:

P(Gar ∩ Pool) = P(Gar) * P(Pool)

0.16 ≠ 0.56*0.27 = 0.15

Then, the events Gar and Pool are not independent

d) Gar and Pool are not mutually exclusive, because there are houses with both pool and garage. We know that because P(A) is not zero.

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3 years ago

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

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Given:

$f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6$

Required:

$We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.$

Explanation:

Given equation is

$F(x)=\sqrt[4]{f(x)g(x)}.$ $F(x)=(f(x)g(x))^{\frac{1}{4}}$ $F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}$

Differentiate the given equation for x.

$Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.$

$F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)$ $F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)$

Replace x=3 in the equation.

$F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)$ $Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}$ $F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)$ $F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-6-1}{4}$ $F^{\prime}(3)=\frac{-7}{4}$

Final answer:

$F^{\prime}(3)=\frac{-7}{4}$

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Answer:

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Step-by-step explanation:

<u><em>There is some data missing in the question, so complete question is below:</em></u>

$Sam\ and\ Lou\ need\ a\ total\ of\ 1\ foot\ of\ wire\ for\ science\ project.$ $Sam's\ wire\ measured\ 8/12\ foot\ long.\ Lou's\ measured\ 7/8-\ foot \ \ long.$ $Do\ they\ have\ enough\ wire\ for\ the\ science\ project?$ $Explain \ your\ reasoning.$

Now, to get that whether they have enough wire for the science project or not.

Measurement of Sam's wire = $\frac{8}{12} \ foot.$

Measurement of Lou's wire = $\frac{7}{8}\ foot.$

Now, adding both the wires of Sam and Lou:

$\frac{8}{12} +\frac{7}{8} \\\\=\frac{16+21}{24} \\\\=\frac{37}{24}$

$=1\frac{13}{24} \ feet.$

<em>As, Sam and Lou need a total of 1 foot of wire for the science project.</em>

<em>And, we see that the Sam's wire and Lou's wire combining gets </em> $1\frac{13}{24}\ feet$ <em> which is</em> <em>more than 1 foot.</em>

<em>Thus, wire they have is enough for science project.</em>

Therefore, yes, Sam and Lou have enough wire for science project.

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